Question: $f(1)=-\frac{1}{2}\,$, $~f\,^\prime(1)=2\,$, $~f\,^{\prime\prime}(1)=1\,$, and $~f\,^{\prime\prime\prime}(1)=-\frac{1}{6}\,$. What are the first four nonzero terms of the Taylor series, centered at $x=1$, of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-\frac{1}{2}+2x+\frac{1}{2}{{x}^{2}}-\frac{1}{36}{{x}^{3}}$. (Choice B) B $-\frac{1}{2}+2x+\frac{1}{2}{{x}^{2}}-{{x}^{3}}$. (Choice C) C $-\frac{1}{2}+2(x-1)+\frac{1}{2}{{(x-1)}^{2}}-\frac{1}{18}{{(x-1)}^{3}}$. (Choice D) D $-\frac{1}{2}+2(x-1)+\frac{1}{2}{{(x-1)}^{2}}-{{(x-1)}^{3}}$. (Choice E) E $-\frac{1}{2}+2(x-1)+\frac{1}{2}{{(x-1)}^{2}}-\frac{1}{36}{{(x-1)}^{3}}$.
Explanation: We know the formula for a Taylor series centered at $~x=1$ for the function $~f\,$. $ f(1)+f\,^\prime(1)(x-1)+\frac{f\,^{\prime\prime}(1)}{2!}{{(x-1)}^{2}}+...+\frac{{{f}^{(n)}}(1)}{n!}{{(x-1)}^{n}}+...$ If we substitute the given function and derivative values, we obtain the following Taylor polynomial. $ T_3(x)=-\frac{1}{2}+2(x-1)+\frac{1}{2!}{{(x-1)}^{2}}+\frac{-\frac{1}{6}}{3!}{{(x-1)}^{3}}$ This simplifies to $ T_3(x)=-\frac{1}{2}+2(x-1)+\frac{1}{2}{{(x-1)}^{2}}-\frac{1}{36}{{(x-1)}^{3}}\,$.